Unit 10- Acids & Bases

14.1- The Nature of Acids & Bases

1. Arrhenius Concept

Acids produce hydrogen ions (protons) H+

Ex. HCl H+ + Cl-

Bases produce hydroxide ions OH-

Ex. NaOH Na+ + OH-

2. Brønsted-Lowry ModelAn acid is a proton (H+) donorA base is a proton (H+) acceptor

3. Hydronium Ion

H3O+

This is how H+ exists in water

4. Conjugate Acid-Base PairConsists of 2 substances related to each other by the donating & accepting of a single proton (H+)

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Acid Base Conj. A Conj. B

5. Acid Dissociation Constant (Ka)Represented by the equilibrium expression (Keq) for an acid

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Why is H2O not included?

Because it is a liquid!

Ex. 14.1Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids:

a) HCl(aq)

b) HC2H3O2(aq)

c) NH4+(aq)

d) C6H5NH3+(aq)

e) [Al(H2O)6]+3(aq)

14.2- Acid Strength

1. A strong acid has a large Ka

Equilibrium lies far to the right & nearly all acid molecules dissociate & donate an H+

Strong acids yield weaker conjugate bases

Strong AcidsHCl, HBr, HI, H2SO4, HNO3, HClO4

2. A weak acid has a small Ka

Equilibrium lies far to the left & only a small amount of acid dissociates & donates an H+

Weak acids yield stronger conjugate basesThe weaker the acid, the stronger its conjugate base

ExamplesHF, HC2H3O2 (acetic acid), H3PO4, H2CO3

3. Monoprotic AcidAn acid having one acidic proton (H+)Ex. HI, HBr, HCl, HF (All Halogens)

4. Polyprotic Acids

a) Diprotic AcidAn acid with 2 acidic protons (H+)Ex. H2SO4, H2CO3

b) Triprotic AcidAn acid with 3 acidic protons (H+)Ex. H3PO4

5. Oxyacids The acidic proton is on an oxygen atomEx. H3PO4, H2SO4, HNO2, etc.

6. Organic AcidsAcids with a carbon backbone & a carboxyl group (-COOH):

Ex. 14.2Using the information below, arrange the following species according to their strengths as bases: H2O, F-, Cl-, NO2

-, CN-

Ka(HF) = 7.2 x 10-4

Ka(HNO2) = 4.0 x 10-4

Ka(HCN) = 6.2 x 10-10

Ka(HCl) = very large

7. H2O

Is amphoteric- can act as an acid or a base

Autoionization of H2O:

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

Keq = [H3O+][OH-]

Kw = [H+][OH-]

Where Kw is the Dissociation Constant for H2O

At 25°C, [H+] = [OH-] = 1.0 x 10-7 M So: Kw = (1.0 x 10-7)(1.0 x 10-7)

Kw = 1.0 x 10-14

The importance of this:In aqueous solutions at 25°C, no matter what it contains, the product of [H+] and [OH-] will always be 1.0 x 10-14

Ex. 14.3Calculate [H+] or [OH-] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic.

a) 1.0 x 10-5 M OH-

b) 1.0 x 10-7 M OH-

c) 10.0 M H+

14.3- The pH Scale1. A scale to measure acidity from 0 140 acidic 7 neutral 14 basic

2. pH Equation:

pH = -log[H+]Example:If [H+] = 1.0 x 10-5

then,

pH = -log(1.0 x 10-5)pH = 5

3. pOH Equation:

pOH = -log[OH-]Measures basicity0 basic 7 neutral 14 acidic

4. Relation to Equilibrium Constant:

pK = -log(K)

Ex. 14.5Calculate the pH and pOH for each of the following solutions at 25°Ca) 1.0 x 10-3M OH-

b) 1.0M H+

Ex. 14.6The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H+], and [OH-] for the sample.

14.4- pH of Strong Acids

1. Strong acids dissociate completely so the concentration of H+ is the same as the concentration of the acid

ExampleIf [HCl] = 0.1MThen [H+] = 0.1MBecause HCl H+ + Cl-

0.1M 0.1M

Ex. 14.7a) Calculate the pH of 0.10M HNO3

b) Calculate the pH of 1.0 x 10-10M HCl

14.5- pH of Weak AcidsThey do not ionize completely

1.Write dissociation reaction

2.Write equilibrium expression

3.List initial concentrations

4.Set up ICE Box in terms of M (set up row E in terms of x)

5.Sub equilibrium concentrations into the equilibrium expression

6.Eliminate x-factor (because Ka is small) and solve for x

7.Calculate [H+] and pH

Example

Calculate the pH of a 1M solution of HF (Ka = 7.2 x 10-4)

Ex. 14.8Calculate the pH of a 0.100M aqueous solution of hypochlorous acid (HOCl, Ka = 3.5 x 10-8)

If 2 weak acids are present, the stronger of the two (the one with the larger Ka) will dominate & control the pH The addition of H+ ions by the weaker acid is insignificant in regards to effecting the pH

Ex. 14.9Calculate the pH of a solution that contains 1.00M HCN (Ka = 6.2 x 10-10) and 5.00M HNO2 (Ka = 4.0 x 10-4)

14.6- Bases

1. Strong BasesCompletely dissociate (ionize) in H2O

Group 1 & 2 Metal HydroxidesKOH, NaOH, LiOH, Mg(OH)2, Ca(OH)2, etc.

2. pH of Strong BasesEx. 14.12

Calculate the pH of a 0.050M NaOH solution.

3. Weaker bases that do not involve a hydroxide ion can create one in H2O:

NH3 + H2O NH4+ + OH-

Weak base

4. Base Dissociation Constant (Kb)

B(aq) + H2O(l) BH+(aq) + OH-(aq)

5. pH of Weak BasesEx. 14.13

Calculate the pH for a 15.0M solution of NH3 (Kb = 1.8 x 10-5)

Ex. 14.14 Calculate the pH of a 1.0M solution of methylamine (CH3NH2).

(Kb = 4.38 x 10-4)

14.7- Polyprotic Acids

1. They dissociate in a stepwise mannerExampleH2CO3(aq) H+(aq) + HCO3

-(aq)

Ka1 = 4.3 x 10-7

HCO3-(aq) H+(aq) + CO3

-2(aq)

Ka2 = 5.6 x 10-11

Ka1 > Ka2

As an H+ is taken, the following H+ is always less acidic

Since the first Ka is larger it will dominate the pH (the amount of H+ given off by the second step is insignificant compared to the amount given off by the first)

Ex. 14.15 Calculate the pH of a 5.0M H3PO4 solution & the equilibrium

concentrations of the species H3PO4, and H2PO4-.

Ex. 14.16 Calculate the pH of a solution 1.0M H2SO4 solution.

14.8- Salts1. Salt- an ionic compound

2. Salts that produce neutral solutionsConsist of the conjugates of strong acids & bases

They have no affinity for, nor do they produce any H+ or OH-

Examples Cl- (from HCl)NO3

- (from HNO3)

Na+ (from NaOH)

3. Salts that produce Basic SolutionsThey are the conjugate bases of weak acidsExample: NaC2H3O2

C2H3O2-(aq) + H2O(l) HC2H3O2(aq) + OH-(aq)

Acts as a base and creates

Crazy Cool Derivation!

Ex. 14.18Calculate the pH of a 0.30M NaF solution. The Ka

value for HF is 7.2 x 10-4.

4. Salts that produce Acidic SolutionsThey are the conjugate acids of weak basesExample: NH4Cl

NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

Acts as an acid and creates

Ex. 14.19Calculate the pH of a 0.10M NH4Cl solution. The Kb

value for NH3 is 1.8 x 10-5.

14.10- The Acid & Base Properties of Oxides

1. Acid OxidesCovalent & form acidic solutionsExamples: SO2, CO2, NO2, etc.

CO2(g) + H2O(l) H2CO3(aq) H+(aq) + HCO3-(aq)

CO2 forms an acidic solution when dissolved in water!

2.Basic OxidesIonic and form basic solutionsExamples: CaO, K2O, etc.

CaO(s) + H2O(l) Ca(OH)2(aq)

Because:CaO(s) Ca+2(aq) + O-2(aq)O-2(aq) + H2O(l) OH-(aq) + OH-(aq)

14.11- The Lewis Model1. Lewis AcidAn electron pain acceptor

2. Lewis BaseAn electron pair donor

Ex 14.22For each reaction, identify the Lewis Acid and Lewis Base

a)Ni+2(aq) + 6NH3(aq) Ni(NH3)6+2(aq)

b)H+(aq) + H2O(l) H3O+(aq)

4.8- Acid-Base Reactions1. Neutralization Reaction:

Acid + Base Salt + Water

2. Titrationsa) Volumetric Analysis- technique for determining

the amount of a certain substance by doing a titration

b) Titration- involves delivery (from a buret) of a measured volume of a solution of known concentration (the titrant) into a solution of the substance being analyzed

c) Equivalence (Stoichiometric) Point- the point where enough titrant has been added to react exactly with the substance being analyzed (the analyte)

d) Indicator- often marked the equivalence point by a change in color (added in the beginning)

e) Endpoint- when indicator changes colorGoal: choose an indicator that matches the endpoint with the equivalence point

Before Titration

During Titration

Equivalence Point

Too Far!!

Ex. 4.12What volume of a 0.100M HCl solution is needed to neutralize 25.0mL of 0.350M NaOH?

Ex. 4.1328.0mL of 0.250M HNO3 and 53.0mL of 0.320M KOH are

mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH- ions in excess after the reaction goes to completion.

Ex. 4.141.3009g of acidic KHC8H4O4 (MW = 204.22 g/mol) is titrated with 41.20mL of NaOH solution until the endpoint is reached. Calculate the concentration of the basic solution used.

Ex. 4.15 A 0.3518g sample of waste containing an unknown amount of

benzoic acid (HC7H5O2, MW = 122.12 g/mol) was titrated with 10.50mL of 0.1546M NaOH solution until endpoint was reached. Calculate the mass percent of HC7H5O2 in the sample.

15.1- Common Ions with Acids & Bases

Use the same procedure as beforeSet up an ICE Box with MolarityPlug in equilibrium concentrations and solve for x

Ex. 15.1 Calculate [H+] in a solution containing 1.0M HF (Ka = 7.2 x 10-4)

and 1.0M NaF.

15.2- Buffered Solutions

1. A buffered solution is one that resists a change in pH

2. Buffer- a weak conjugate acid-base pairA weak acid and its conjugate base OR a weak base and its conjugate acid

ExamplesHF & NaFNH3 & NH4Cl

Think of a buffer like a sponge that absorbs H+ and OH- so that the pH doesn’t change too much

Example: Weak Acid/Conjugate Base Buffer

Initial Ratio: [HA] / [A-] = 0.5 / 0.5Then 0.01 mol OH- is added and changes HA to A- (if H+ is added it will change A- HA)

Final Ratio: [HA] / [A-] = 0.49 / 0.51The OH- added changes HA to A- If H+ is added it will change A- HA

Ex. 15.2A buffered solution contains 0.50M acetic acid (HC2H3O2,

Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Calculate the pH of the solution.

Ex. 15.3 Calculate the change in pH that occurs when 0.010 mol solid NaOH

is added to 1.0 L of the buffered solution described in Ex. 15.2. Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0L of water.

2. Henderson-Hasselbalch Equation

OR

Ex. 15.4Calculate the pH of a solution containing 0.75M lactic

acid (Ka = 1.4 x 10-4) and 0.25M sodium lactate.

Ex. 15.5A buffered solution contains 0.25M NH3 (Kb = 1.8 x 10-5)

and 0.40M NH4Cl. Calculate the pH of this solution.

Ex. 15.6Calculate the pH of the solution that results when 0.10mol

gaseous HCl is added to 1.0L of the buffered solution from Ex. 15.5.

15.3- Buffering Capacity

Represents the amount of H+ or OH- the buffer can absorb without a significant change in pH

The pH of a buffered solution is dependent on [A-] / [HA]

When [A-] = [HA]:pH = pKa + log (1)

pH = pKa

and THIS is the most effective buffer!

Ex. 15.7 Calculate the change in pH that occurs when 0.010mol gaseous

HCl is added to 1.0L of each of the following solutions: Solution A: 5.00M HC2H3O2 & 5.00M NaC2H3O2

Solution B: 0.050M HC2H3O2 & 0.050M NaC2H3O2

For acetic aced, Ka = 1.8 x 10-5

One should choose a buffer with a pKa closest to the desired pH

Ex. 15.8A chemist needs a solution buffered at pH 4.30 and can choose from the following acids and their sodium salts. Which choice should be made?a) Chloroacetic acid (Ka = 1.35 x 10-3)

b) Propanoic acid (Ka = 1.3 x 10-5)

c) Benzoic acid (Ka = 6.4 x 10-5)

d) Hypochlorous acid (Ka = 3.5 x 10-8)

15.4- Titrations and pH Curve1. pH (Titration) Curve

A plot of solution pH vs. Volume of titrant added during a titration

2. Strong Acid – Strong Base TitrationThe Net Ionic Reaction is always:

H+(aq) + OH-(aq) H2O(l)

Small amounts are used sometimes, so we can use mmol1 mmol = 10-3 mol0.005 mol = 5 mmolM = mmol / mL

Example

50.0mL of 0.2M HNO3 titrated with 0.1M NaOH

Solve for the pH at Various Points of the Titration

A. No NaOH has been added:

B. 10mL of 0.1M NaOH added:

C.100mL 0.1M NaOH added: (Equivalence Point)

The pH at the equivalence point for a strong acid – strong base titration is always 7!

D. 150mL of 0.1M NaOH added:

pH will just keep increasing

Graph: Strong Acid – Strong Base Titration

3. Weak Acid – Strong Base Titration

Example: 50.0mL of 1.0M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) with 0.1M NaOH

Solve for the pH at Various Points of the Titration

A. No NaOH has been added:

B. 10mL of 0.1M NaOH added:

C. 25mL 0.1M NaOH added: (Halfway Point)

The pH will always equal the pKa at the Halfway Point

D. 50mL 0.1M NaOH added: (Equivalence Point)

We must find the pH using the Kb of C2H3O2- :

The pH at the equivalence point of a weak acid – strong base titration is always greater than 7.

E. 60mL 0.1M NaOH added:

Graph: Weak Acid – Strong Base Titration

4. Weak Base – Strong Acid Titration

Same procedure as for weak acid – strong base titration, just opposite

The pH of the solution at the Halfway Point is equal to the pKb

The pH of the solution at the Equivalence point is always less than 7.

Graph: Weak Base – Strong Acid Titration

Ex. 15.9Hydrogen cyanide gas (HCN, Ka = 6.2 x 10-10) is dissolved

in water. If a 50.0mL sample of 0.10M HCN is titrated with 0.10M NaOH, calculate the pH of the solution:

a) After 8.00mL of 0.10M NaOH is added

b) At the halfway point of the titration

c) At the equivalence point of the titration

15.5- Indicators

1. Two methods for determining the equivalence point:

a) pH Meter- monitor pH & plot the titration curve

b) Indicator- changes color at specific pH’s

2. IndicatorsComplex molecules that are actually themselves weak acids

HIn(aq) H+(aq) + In-(aq)

(Color 1) (Color 2)Ex. Phenolphthalein

HPh(aq) H+(aq) + Ph-(aq)

(colorless) (pink)Color change will be sharp, occurring with the addition of a single drop of titrant

pH Color

Less than 8 colorless

Between 8 & 9 light pink

Greater than 9 pink

Finally done with Unit 10!